1
Chapter 7 The Transcendental Functions
7
The Transcendental Functions
7.1 The Natural Logarithmic Function
1.
a. ln 6  ln(2  3)  ln 2  ln 3  0.6931  1.0986  1.7917
b. ln 32  ln 3  ln 2  1.0986  0.6931  0.4055
2.
a. ln 30  ln(2  3  5)  ln 2  ln 3  ln 5  0.6931  1.0986  1.6094  3.4011
b. ln 7.5  ln 152  ln 3  ln 5  ln 2  1.9086  1.6094  0.6931  2.0149
3. ln 2 5 3  ln  2  31/2  51   ln 2  12 ln 3  ln 5
4. ln x z1/2y  ln  x1/3 y 2/3 z 1/2   13 ln x  32 ln y  12 ln z
1/3 2/3
 x 1 
5. ln 

 x 1 
1/3
1 x 1 1
 ln
 3 ln( x  1)  13 ln( x  1)
3 x 1
6. ln 4  ln 6  ln12  ln 4126  ln 2
7. 3ln 2  12 ln( x  1)  ln 23  ln( x  1) 1/2  ln
8.
9.
8
x 1
2
Chapter 7 The Transcendental Functions
10.
11. f ( x)  ln(2 x  1) . We require that 2 x  1  0  x   12 , so D  ( 12 , ) .
12. g ( x)  ln cos x . We require that cos x  0  2k  2  x  2k  2 for
k  0, 1, 2,... .
13. f ( x) 
d
2
ln(2 x  3) 
dx
2x  3
14. h( x)  ln x  12 ln x  h( x) 
15. g (u )  ln
1 d
1
ln x 
2 dx
2x
u
1
1
1
 ln u  ln(u  1)  g (u )  

u 1
u u  1 u (u  1)
16.
y  x(ln x) 2  y  x
d
d
1
(ln x) 2  (ln x) 2  ( x)  x  2 ln x   (ln x) 2  (ln x) 2  2 ln x
dx
dx
x
d ln x
17. g ( x) 

dx x  1
1
( x  1)  ln x
x  1  x ln x x(1  ln x)  1
x


2
( x  1)
x( x  1)2
x( x  1)2
d
(ln x)
d
1
dx
18. f ( x)  [ln(ln x)] 

dx
ln x
x ln x
19. f ( x)  ln( x ln x)  ln x  ln(ln x) , so using the result of Exercise 37,
f ( x) 
1
1
ln x  1


.
x x ln x
x ln x
3
Chapter 7 The Transcendental Functions
20. g ( x) 
d
d
cos(ln x)
[sin(ln x)]  cos(ln x)  ln x 
dx
dx
x
21. f ( x) 
d 2
d
 sin x
( x ln cos x)  2 x ln cos x  x 2 ln cos x  2 x ln cos x  x 2 
dx
dx
cos x
 2 x ln cos x  x 2 tan x
22. h(u ) 
d
sec u tan u
ln | sec u |
 tan u
du
sec u
23. g (t )  ln
g (t ) 
sin t  1
 ln | sin t  1|  ln | cos t  2 |
cos t  2
cos t
( sin t )
2cos t  sin t  1


sin t  1 cos t  2 (sin t  1)(cos t  2)
24. ln y  x ln x  1 
25. ln
1
1

 y   ln x  x    0  y  y (ln x  1)
y
x

x
 x  y 2  0  ln x  ln y  x  y 2  0 . Differentiating, we obtain
y
1

1 y
1
  1  2 yy  0  y   2 y   1 
x y
x
y

26. y  2 x 2  3x 2 ln x  y  4 x  6 x ln x  3 x 2 
y  7  6 ln x  6 x 
1
 7 x  6 x ln x 
x
1
 13  6 ln x . Substituting into the differential equation gives
x
x 2 y  3xy  4 y  x 2 (13  6ln x)  3x(7 x  6 x ln x)  4(2 x 2  3x 2 ln x)  0 , so the
equation is satisfied, and y  2 x 2  3x 2 ln x is indeed a solution.
1
 ln x  1  y  |x 1  1 , the slope of the required tangent
x
line. An equation of the line is y  0  1( x  1)  y  x  1 .
27. y  x ln x  y  ln x  x 
28.
f is continuous on the closed interval
 12 , 2 .
f ( x)  x ln x  x  f ( x)  ln x  x(1/ x)  1 
ln x  0  x  1 is a critical number of f in
1
2
f ( x)
 0.85
1
0.61
(10)
(2) There is no
y-intercept. The x-intercept lies between 0 and 1.
(3) There is no symmetry.
1
2
 12 , 2 .
29.
f ( x)  x  ln x
(1) The domain of f is (0, ) .
x
4
Chapter 7 The Transcendental Functions
(4) lim f ( x)  lim( x  ln x)  
x 
x 
(5) lim f ( x)  lim ( x  ln x)   , so x  0 is a
x 0
x 0
vertical asymptote.
(6) f ( x)  1  1/ x  0 on (0, ) , so f is increasing on (0, ) and has no critical
number. (7) f has no relative extremum. (8) f ( x)  1/ x 2  0 on (0, ) , so f
is concave downward on (0, ) . (9) f has no inflection point.
30.
f ( x)  ln( x 2  1)
(1) The domain of f is (, ) .
(2) The x- and
y-intercepts are 0.
(3) f ( x)  ln  ( x) 2  1)   ln  x 2  1  f ( x) , so
(10)
f is symmetric with respect to the y-axis.
(4) lim f ( x)  lim ln( x2  1)   and
x 
x 
lim f ( x)   by symmetry.
x 
(5) There is no
asymptote.
2x
 0  x  0 , so 0 is a critical number. From the sign diagram, we
x 1
see that f is decreasing on (, 0) and increasing on (0, ) . (7) f (0)  0 is a
(6) f ( x) 
2
( x 2  1)(2)  2 x(2 x) 2(1  x 2 )
 2
 0  x  1 .
( x 2  1)2
( x  1)2
From the sign diagram, we see that f is concave downward on (, 1)  (1, ) and
concave upward on ( 1,1) . (9) f has inflection points at (1, ln 2) and (1, ln 2) .
relative minimum value. (8) f ( x) 
31. Let f ( x)  x ln x  1 . Then f ( x)  ln x  x(1/ x)  ln x  1 . The Newton iteration
formula is xn 1  xn 
f ( xn )
x ln xn  1
x 1
 xn  n
 n
. Taking x0  2 , we have
f ( xn )
ln xn  1
ln xn  1
x1  1.771848 , x2  1.763236 , x3  1.763223  x4 , so the root is approximately
1.76322.
5
Chapter 7 The Transcendental Functions
32.
y  (2 x  1) 2 (3 x 2  4)3  ln y  2 ln(2 x  1)  3ln(3 x 2  4)
y
22
3 6x
2(24 x 2  9 x  8)

 2

y 2 x  1 3 x  4 (2 x  1)(3 x 2  4)

 y  2(2 x  1)(3 x 2  4) 2 (24 x 2  9 x  8)
33.
x 1
1 x 1 1
 ln y  ln 2
 3 [ln( x  1)  ln( x 2  1)]
2
x 1
3 x 1
y 1 1
2x
x2  2x 1
  (
 2 )
y 3 x 1 x 1
3( x  1)( x 2  1)
y
3
 y  
x2  2 x  1
3( x  1) 2/3 ( x 2  1) 4/3
34. Taking logarithms of both sides gives ln y  ln x x  x ln x , so
y
 x( 1x )  ln x  1  ln x  y  (ln x  1) x x Therefore,
y
y  (ln x  1) dxd x x  x x
35.
2
 3x dx 
2
3
d
dx
(ln x  1)  (ln x  1)(ln x  1) x x  x x ( 1x )  [ x(ln x  1) 2  1]x x 1 .
ln | x | C
36. Let u  3 x 3  1 . Then du  9 x 2 dx , x  0  u  1 , and x  1  u  4 . Thus,
x2
1 4 du 1
4
dx

 9 ln| u | 1  19 ln 4  92 ln 2 .
0 3x3  1

1
9 u
1
37. Let u  x1/3  1 . Then du  13 x 2/3dx , so
x
2/3
1
du
dx  3
 3ln | u | C  3ln | x1/3  1| C .
1/3
( x  1)
u
38. Let u  ln x . Then du 
dx
, so
x
1
 x ln x dx  
du
 ln | u | C  ln | ln x | C .
u
39. Let u  1  sin x , Then du  cos xdx , so
cos x
 1  sin x dx  
du
 ln | u | C  ln |1  sin x | C .
u
40.
sec   cos  sec   tan 
sec2   sec  tan 

d  
d   cos  d
 (sec  cos )d  
1
sec   tan 
sec   tan 
. To evaluate the first integral, we let u  sec  tan  . Then
du  (sec tan   sec2  )d , so
6
Chapter 7 The Transcendental Functions
du
 cos  d  ln | u |  sin   C  ln | sec   tan  |  sin   C
u 
41. Let u  2  x ln x . Then du  (ln x  1)dx ,
 (sec  tan  )d  
so 
1  ln x
du
dx  
 ln | u | C  ln | 2  x ln x | C .
2  x ln x
u
42.
0
1
A   ( x2  x2x1 )dx   ( x2x1  x2 )dx  [ 16 x3  12 ln( x 2  1)]01  [ 12 ln( x 2  1)  16 x3 ]10
2
1
2
0
  16  12 ln 2  12 ln 2  16  ln 2
43.


y  12  x x 2  1  ln x  x 2  1  


x 

1
2


2
x
x  1   x  1  x 2  1 , so
y  12  x 2  1  x 


x2 1 x  x2 1 
x2 1


1  ( y)2  1  ( x 2  1)2  x 2 and L  
3
1
3
1  ( y)2 dx   x dx  12 x 2  4 .
3
1
1
44.
 c ln t dt  x ln t dt   d   x ln t dt  x ln t dt 
x
c
c
 x
 dx  c

  ln x  (ln x 2 ) dxd ( x 2 )   ln x  2 x ln x 2  (4 x  1) ln x
dy
dx

d
dx
x2
ln t dt 
2
2
d
dx
20
20
0
0
45. The distance traveled is S   v(t )dt  
1056t
dt . Let u  t 2  36 . Then
t 2  36
du  2t dt , t  0  u  36 , and t  20  u  436 . Thus,
S
46.
a.
1056 436 du
436
 528ln u |36
 528(ln 436  ln 36)  1317 ft.

36
2
u
7
Chapter 7 The Transcendental Functions
v0
dx d  1
dv d  v0 
 1 v0 k
  ln(v0 kt  1)  

a
 

dt dt  k
dt dt  v0 kt  1 
 k v0 kt  1 v0 kt  1
kv02
d
 v0 (v0 kt  1) 1  v0 (1)(v0 kt  1) 2 (v0 k )  
dt
(v0 kt  1) 2
v
b. From part a, we see that a   kv 2 , and the result follows.
v0
1
c. From part a, v 
, But x  ln(v0 kt  1)  v0 kt  1  e kx , so
k
v0 kt  1
v
v0
 v0 e  kx .
e kx
47. Differentiating the given equation implicitly gives C
y
Ax
 Dy 
 Bx 
y
x

 Dy  C 
C
( A  Bx) yx
A

 A  Bx  
.
 D   y     B  x  
 y  
 x  y 
y
( Dy  C ) x
x

 x 

 y 
48.
r2
r2 
r r (T  T )  1 1  
r r (T  T ) 
1
1 
r 
Tav 
T1  1 2 2 1    dr 
T1r  1 2 2 1  ln r   



r2  r1 r1 
r1  r2  r r1  
r2  r1 
r1  r2 
r1   r
1


r1r2 (T2  T1 ) 
r2   
r1r2 (T2  T1 )
1  
 ln r1  1 
 T1r2 
 ln r2     T1r1 
r2  r1  
r1  r2 
r1   
r1  r2
 


r1r2 (T2  T1 ) r2
r r (T  T ) r r (T  T )
1 
ln  r2 (T2  T1 )   T1  1 2 1 22 ln 2  2 2 1
(r2  r1 )T1 
r2  r1 
r1  r2
r1
(r2  r1 )
r1
r2  r1

49. False. Take a  2 and b  1 . Then ln a  ln b  ln 2  ln1  ln 2 , but
ln(a  b)  ln(2  1)  ln1  0 , so ln a  ln b  ln(a  b) .
if x  0
ln x
50. True. If x  0 , then f ( x)  ln | x | 
ln( x) if x  0
51. True. g ( x)  ln x is continuous on (1, ) and g ( x)  0 on (1, ) , so
f ( x) 
1
1
is continuous on (1, ) .

g ( x) ln x
52. False. The integrand is not defined at x  2 , so neither integral is defined.
8
Chapter 7 The Transcendental Functions
7.2 Inverse Functions
1. f ( g ( x))  f ( 3 3x )  13 ( 3 3x )3  13 (3x)  x and g ( f ( x))  g ( 13 x3 )  3 3( 13 x3 )  x .
(2 x  3)  3
 x 3
 x 3
 x.
2. f ( g ( x))  f 
  2
  3  x and g ( f ( x))  g (2 x  3) 
2
 2 
 2 
3. f ( g ( x))  f ( 18 ( x3/2  8))  4( 18 ( x3/2  8)  1) 2/3  4( 18 x3/2 )  x and
g ( f ( x))  g (4( x  1) 2/3 )  18 {[4( x  1) 2/3 ]3/2  8}  18 [8( x  1)  8]  x .
4. f is one-to-one since there is no horizontal line that cuts the graph of f at more than
one point.
5. f is not one-to-one. The horizontal line shown cuts the graph of f at infinitely many
points.
6. f is not one-to-one because there is at least one horizontal line that cuts the graph of
f at infinitely many points.
7. f is one-to-one. Any horizontal line cuts the graph of f at exactly one point.
8. f ( x)  1  x is one-to-one. There is no horizontal line that cuts the graph of f at
more than one point.
9. f ( x)   x 4  16 is not one-to-one. The horizontal line y  5 cuts the graph of f
at two points.
9
Chapter 7 The Transcendental Functions
10. By the definition of inverse functions, f 1 (5)  2 .
11. By inspection f (0)  1 , so f 1 (1)  0 .
12. By inspection f ( 6 )  1 , so f 1 ( 1)  6 .
13. By inspection f ( 4 )  cot 2  0 , so f 1 (0)  4 .
14.
15. Put y  3x  2 . Then x  13 ( y  2) , so f 1 ( x) 
x2
.
3
16. Put y  x3  1 . Then x  3 y 1 , so f 1 ( x)  3 x  1 .
10
Chapter 7 The Transcendental Functions
17. Put y  9  x 2 , x  0 . Then x  9  y 2 , so f 1 ( x)  9  x 2 , x  0 .
18.
a. F  95 C  32  95 C  F  32  C  95 ( F  32) , and so C  f 1 ( F )  95 ( F  32) .
This formula gives the temperature C in degrees Celsius as a function of the
temperature F in degrees Fahrenheit.
b. We require that 59 95 ( F  32)  273.15  F  32  95 (273.15)  459.67 , so the
domain of f 1 is [459.67,  ) .
19.
a. Put y  10.72(0.9t  10)3/10 . Then
10/3
 y 
0.9t  10  

 10.72 
10  y 
 t  

9  10.72 
10/3
10  25 
b. f (25)  

9  10.72 
10/3
1
10/3


10  p 
1
 10  , so f ( p)  
  10  .
9  10.72 



 10   7.58 , so the proportion of of Americans aged 55

and over will be 25% around the middle of 2007.
20.
a. f ( x)  x 2 on [0, )  g ( x)  f 1 ( x)  x , Using Formula 3,
11
Chapter 7 The Transcendental Functions
g ( x) 
1
1

f ( g ( x)) 2 y

y x
1
2 x
.
b. From the result of part a, g ( x)  x , so g ( x) 
1
2 x
.
21.
a. f ( x)  2x  1  f (2)  2  2  1  5 , so (2,5) lies on the graph of f .
b. g (5) 
1
1
1


f ( g (5)) f (2) 2
22.
a. f ( x)  x5  2x3  x  1  f (0)  1 , so (0, 1) lies on the graph of f .
b. g (1) 
1
1
1

 4
1
f ( g (1)) f (0) 5x  6 x 2  1 x0
23.
a. f ( x)  ( x3  1)3  f (1)  (1  1) 3  8 , so (1,8) lies on the graph of f .
b. f ( x)  3( x3  1)2 (3x2 )  9 x2 ( x3  1)2 
g (8) 
1
1
1

 2 3
f ( g (8)) f (1) 9 x ( x  1) 2

x 1
1
36
24.
a. f ( x) 
1
 f (2)  15 , so (2, 15 ) lies on the graph of f .
1  x2
b. f ( x)  
25. g (4) 
2x
1
1
(1  x 2 )2
1


g
(
)




5
(1  x 2 )2
f ( g ( 15 )) f (2)
2x

x 2
25
4
1
1
1


f ( g (4)) f (2) 3
26. Observe that f (2)  
2
2
dt
1 t3
 0 , showing that (2, 0) lies on the graph of f.
d x dt
1

(by the FTC, Part 1). Therefore,

dx 2 1  t 3
1  x3
1
1
( f 1 )(0) 

 1  23  3
1


f ( f (0)) f (2)
Next, observe that f ( x) 
27. Suppose f has an inverse. If f ( x1 )  f ( x2 ) , then f 1 ( f ( x1 ))  f 1 ( f ( x2 ) ), so
x1  x2 . This shows that f is one-to-one. Next, suppose that f is one-to-one. Then for
each number b in the range of f , there is exactly one number a in its domain such that
f (a )  b . Define the function g, whose domain is the range of f , by the rule
12
Chapter 7 The Transcendental Functions
g (b)  a for any b in its domain. Then g  f 1 .
28. True. This follows from the definition of the inverse of a function.
29. True. f ( x)  1/ x 2 is one-to-one on (, 0) as well as on (0, ) , so if (a, b) is
an interval not containing 0, then f is one-to-one and has an inverse.
30. True. See Theorem 2.
31. True. f ( x)  (2n  1)a2n 1 x 2n  (2n  1)a2n 1 x 2n 2  (2n  3)a2 x 2n 4  ...  a1  0  f
is monotonically increasing  f  exists.
13
Chapter 7 The Transcendental Functions
7.3 Exponential Functions
1.
a. ln e3  3ln e  3
b. ln e x  x 2 ln e  x 2
2.
2
a. e2ln3  eln3  32  9
2
b. eln
x
 x
3.
a. eln x  2  x  2
b. ln e2 x  3  2 x  3  x   32
4.
a. 2e x  2  5  e x  2  52  x  2  ln 52  x  ln 52  2
b. ln x  1  1  x  1  e  x  e2  1
5.
a.
50
 20  1  4e0.2 x 
1  4e0.2 x
50
20
 4e0.2 x  23  e0.2 x  83  0.2 x  ln 83  5ln 83
b. e2 x  5e x  6  0 . Let u  e x . Then we have
u 2  5u  6  (u  3)(u  2)  0  u  2 or u  3  e x  2 or e x  3  x  ln 2 or
x  ln 3 .
6. f ( g ( x))  f (ln x )  e2ln( x
1/2
)
 eln x  x and
g ( f ( x))  g (e2 x )  ln e2 x  ln e x  x .
7. f ( g ( x))  f (2ln x)  e(2ln x )/2  eln x  x and
g ( f ( x))  g (e x /2 )  2 ln e x /2  2  2x  x .
14
Chapter 7 The Transcendental Functions
2e x  1
2  e x
2

lim

x

x
x  3e  2
x  3  2e
3
8. lim
 3t 2  1

9. lim  2 e0.1t   32  0  0
t  2t  1


2e tan x
2etan x
10. lim 
is negative for 0  x  2 and e tan x   ,
  since
x ( /2) 2 x  
2x  
while 2x    0 as x  ( 2 ) .
11. f ( x) 
d 4 x
(e )  4e 4 x
dx
12. f (t ) 
d t
et
t d 1/2
t 1 1/2
e e
t e 2t

dt
dt
2 t
13. g ( x) 
d  e2 x  (1  e x )(2e2 x )  e2 x (e x ) 2e2 x  2e x  e x e x (2e x  3)




dx  1  e x 
(1  e x )2
(1  e x )2
(1  e x )2
14.
d x  x 1/2 1 x  x 1/2 d x  x
(e  e )  2 (e  e )
(e  e )
dx
dx
e x  e x
x
 x 1/2
x
x
1
 2 (e  e ) (e  e ) 
2 e x  e x
f ( x) 
15. y 
d cos x
(e )  ecos x ( sin x)  ecos x sin x
dx
16. h(t ) 
d t ln t

 1 
e  et ln t ln t  t     (1  ln t )et ln t
dx
 t 

17. f ( x) 
18. y 
19.
d
cos e 2 x  ( sin e 2 x )(2e 2 x )  2e 2 x sin e 2 x
dx
d x
2 3

(e  ln x 2 )3  3(e x  ln x 2 ) 2  e x    ( xe x  2)(e x  ln x 2 ) 2
dx
x x

15
Chapter 7 The Transcendental Functions
d 2
2e2 x
[ x ln(e 2 x  1)]  2 x ln(e 2 x  1)  x 2  2 x
dx
e 1
2x
2x
2x
2 x[(e  1) ln(e  1)  xe ]

e2 x  1
f ( x) 
20.
d 2x
1

(e  ln 3x)3/2  32 (e 2 x  ln 3x)1/2  2e 2 x  
dx
x

3

(2 xe2 x  1)(e2 x  ln 3 x)1/2
2x
f ( x) 
21. xe2 y  x3  2 y  5  e2 y  xe2 y (2 y)  3x 2  2 y  0  y 
3x 2  e2 y
2( xe2 y  1)
22.
e x sec y  xy 2  0  e x sec y  (e x sec y tan y ) y  y 2  2 xyy  0 
y 
y 2  e x sec y
e x sec y tan y  2 xy
23. y  2e x /2  5e3 x /2  y  e x /2  152 e3 x /2  y  12 e  x /2  454 e3 x /2 . Substituting into
the differential equation, we have
4 y  4 y  3 y  4( 12 e x /2  454 e3 x /2 )  4(e  x /2  152 e3 x /2 )  3(2e  x /2  5e3 x /2 )  0 . The
differential equation is satisfied, and so y  2e x /2  5e3x /2 is a solution.
24.
y  e x (cos 4 x  2sin 4 x) 
y  e x (cos 4 x  2sin 4 x)  e x (4sin 4 x  8cos 4 x)  e x (9cos 4 x  2sin 4 x) 
y  e x (9cos 4 x  2sin 4 x)  e x (36sin 4 x  8cos 4 x)  e x (cos 4 x  38sin 4 x) .
Substituting, we get
y  2 y  17 y  e x (cos 4 x  38sin 4 x)  2e x (9cos 4 x  2sin 4 x)  17e x (cos 4 x  2sin 4 x)  0
, showing that the given function is a solution of the differential equation.
25. y  xe x  y  e x  xe x  (1  x)e x  y 1  0 . Thus, the slope of the required
tangent line is m  0 , and an equation of the line is y  e1  0( x  1) or y  1/ e .
26.
f ( x)  xe x  f ( x)  e x  xe x  (1  x)e x  0  x  1
. From the table, we see that f has absolute maximum
value f (1)  1/ e and absolute minimum value
f (1)  e .
27.
x
f ( x)
1
1
2
e
1/ e
2 / e2
16
Chapter 7 The Transcendental Functions
f ( x)  xe x
(1) The domain of f is (, ) .
y-intercepts are 0.
(2) The x- and
(3) There is no symmetry.
x
  and lim f ( x)  0 .
x 
x  e x
(4) lim f ( x)  lim
x 
(10)
(5) y = 0 is a horizontal asymptote.
(6) f ( x)  e x  xe x  (1  x)e x  0  x  1 , so
1 is a critical number of f . From the sign
diagram, f is increasing on ( ,1) and
decreasing on (1,  ) . (7) f (1)  e1 is a relative
maximum value.
(8) f ( x)  e x  (1  x)e x  ( x  2)e x  0  x  2 . From the sign diagram, we see
that f is concave downward on ( , 2) and concave upward on (2, ) . (9) f has
an inflection point at. (2, 2e2 ) .
28.
e x  e x
2
(1) The domain of f is (, ) .
f ( x) 
(2) The x- and y-intercepts are 0.
(10)
e x  e x
e x  e x
  f ( x) , so f is

2
2
symmetric with respect to the origin.
(3) f ( x) 
(4) lim f ( x)   and lim f ( x)   .
x 
x 
(5) There is no asymptote.
(6) f ( x)  12 (e x  e  x )  0 for all x in (, ) , so f is increasing on (, ) .
(7)
f has no relative extremum. (8) f ( x)  12 (e x  e  x )  0  e x  e x  0 
e 2 x  1  0  x  0 . From the sign diagram, we see that f is concave downward on
(, 0) and concave upward on (0, ) . (9) f has an inflection point at (0, 0) .
29.
a. The population at the beginning of 2000 was P (0)  0.07 , representing 70,000.
The population at the beginning of 2030 will be P(3)  0.3537 , representing
approximately 353,700.
17
Chapter 7 The Transcendental Functions
b. P(t )  0.0378e0.54t . The population was changing at the rate of P(0)  0.0378 ,
representing 37,800 per decade, at the beginning of 2000. At the beginning of 2030, it
will be changing at the rate of approximately P(3)  0.191 , representing 191,000 per
decade.
30.
a. N (0) 
5000
5000

 50 students
0
1  99e
100
b. N (t ) 
396,000e0.8t
396,000e1.6


N
(2)

 182 students/day
(1  99e0.8t )2
(1  99e1.6 )2
c.
(1  99e0.8t ) 2 e 0.8t (0.8)  e 0.8t (2)(1  99e 0.8t )[99e 0.8t ( 0.8)]
N (t )  396, 000 
(1  99e 0.8t ) 4
 396, 000 

(1  99e0.8t )e 0.8t (0.8)(1  99e 0.8t )
(1  99e 0.8t ) 4
396, 000(0.8e0.8t )(99e0.8t  1)
0
(1  99e 0.8t )3
99e 0.8t  1  0  e 0.8t 
1
1
1
t 
ln
 5.74 , so the flu is spreading most
99
0.8 99
rapidly after about 5.7 days.
5000
 5000 students.
t  1  99e 0.8t
d. lim N (t )  lim
t 
31.
2
2
d
(5.3e0.095t 0.85t )  5.3(0.19t  0.85)e0.095t 0.85t  0  0.19t  0.85  0
dt
 t  4.47 , so N is decreasing on (0, 4) .
a. N (t ) 
b. N (0)  5.3(0.85)e0  4.505 , so the number is being reduced by 4505 cases per
year at the beginning of 1959. N (3)  5.3(0.19  3  0.85)e0.09590.853  0.272 , so the
number is being reduced by 272 cases per year at the beginning of 1962.
32.
F ( x) 
d  10,890e0.1x 
dx  (1  100e0.2 x )3/2 
 (1  100e0.2 x )3/2 (0.1e0.1x )  e0.1x ( 32 )(1  100e0.2 x )1/2 (100e0.2 x )(0.2) 
=10,890 

(1  100e0.2 x )3


0.1 x
0.2 x
1089e (1  200e )

 0  1  200e0.2 x  0  x  26.5.
(1  100e0.2 x )5/2
x
F ( x)
200
2.2 10
5
26.5
50
419.2
4.9  10 4
18
Chapter 7 The Transcendental Functions
From the table, we see that the maximum force is about 419.2 lb.
33.
a. The number of deaths at the beginning of 1950 was N (0)  130.7  50  180.7 per
100,000 people.
b. N (t )  130.7(0.1155)(2t )e0.1155t  30.19te0.1155t
2
2
. The rates of change of the
number of deaths per 100,000 people are as follows:
Year 1950 1960 1970 1980
Rate
0
27
38
32
c. N (t )  30.1917[e0.1155t  t (0.1155)(2t )e0.1155t ]  30.1917(1  0.231t 2 )e0.1155t .
2
2
2
Setting N (t )  0 gives t   2.08 , so t  2 gives an inflection point, and we
conclude that the decline was greatest around 1970.
d. The number is given by N (6)  52 per 100,000 people.
34. We are given that
c(1  e  at /V )  m  1  e  at /V 
m
m
m
 e  at /V   1  e  at /V  1  . Taking
c
c
c
logarithms of both sides of the inequality, we have

at
cm
at
cm
V cm
V
cm V
c
ln e  ln
   ln
 t  ln
 t    ln
  ln
V
c
V
c
a
c
a
c  a cm
. Therefore, the liquid must not be allowed to enter the organ for longer than
t
V
c
ln
minutes.
a cm
35.
a. p(t ) 
d  4(e4t 1) 
d  e 4t  1 
(2e 4t 1)4e 4t  (e 4t 1)(8e 4t )
16e 4t
.

4

4



dt  2e4t  1 
dt  2e4t 1 
(2e4t 1) 2
(2e4t 1) 2
Since p(t )  0 for all t  0 , we see that p is increasing on (0, ) .
1

 1  e 4t
4(e4t  1)
b. lim p(t )  lim
 4lim 
t 
t  2e 4t  1
t 
 2  14t
e


 4
   2 , so in the long run the
 2

concentration p (t) of the product is 2 mol/L.
36.
a. Let t denote the time it takes for the chain to slide off the table. Then
s(t )  12 (e
g /6t
 e
g /6t
)6e
g /6t
 e
g /6t
 12  e2
g /6t
 12e
g /6t
 1  0 . Using the
19
Chapter 7 The Transcendental Functions
quadratic formula, we find e
g /6t

12  144  4
 6  35 or 6  35 . Since
2
ln(6  35) is negative, the second root is rejected, and we have
ln(6  35)
 1.94 s.
g/6
g / 6t  ln(6  35)  t 
b. v(t )  s(t )  12 ( g / 6e
v(1.94)  12 [ 9.8 / 6e
9.8/6 (1.94)
g /6t
 g / 6e
 9.8 / 6e
g /6t
) , so
 9.8/6 (1.94)
]  7.57 m/s .
37.
a. The number was P(10) 
74
1  2.6e
1.66  4.536  6.6
 69.63% .
b. The growth rate would be
P(10) 
74(2.6)(0.166  0.09072t  0.0198t 2 )e0.166t 0.04536t
(1  2.6e0.166t 0.04536t
2
2
0.0066t 3
 5.09% / decade
 0.0066t 3 2
)
t 10
tn  10e0.1tn  41.25
.
1  e0.1tn
Taking t0  40 , we find t1  41.08675 and t2  41.08569  t3 , so t  41.08569 .
38. Let f (t )  t  10e0.1t  41.25 . Use the iteration tn 1  tn 
39. Let u  (ln x) 2  du 
Then

e
ln x
1 x
2ln x
x
dx 
ln x
x
1
1
0
0
dx  12 du , x  1  u  0 , and x  e  u  1 .
e(ln x ) dx  12  eu du  12 eu  12 (e  1) .
2
1
1
1
1
0
0
0
0
40. V    y 2 dx    (e x )2 dx    e2 x dx  2 e2 x  2 (e2  1)
41. False. F (0) 
cos 0 1
 1
e0
1
20
Chapter 7 The Transcendental Functions
42. False.
d x
e  ex
dx
43. True.
e
ln x
dx   x dx  12 x 2  C  12 x  x  C  12 xeln x  C .
21
Chapter 7 The Transcendental Functions
7.4 General Exponential and Logarithmic Functions
1. 2
3
 eln 2  e
3
2. 2tan x  eln 2
tan x
3 ln 2
 etan x ln 2
3. log10 100  log10 102  2 log10 10  2
4. log125 25  log125 1252/3  23 log125 125 
5.
a. 34  81  log 3 81  4
b. 53 
1
1
 log 5
 3
125
125
6.
a. log
1
1
 3  103 
1000
1000
b. log1/3 9  2  ( 13 ) 2  9
7.
8.
2
3
22
Chapter 7 The Transcendental Functions
9. a. log 3 6 
ln 6
 1.631
ln 3
b. log 2 8 
log8
3
log 2
c. log 2  
ln 
 3.303
ln 2
10. f ( x)  3x  f ( x)  (ln 3)3x
11.
 1
f (v)  (cos v)71/ v  f (v)  (  sin v)(71/ v )  cos v(ln 7)(71/ v )   2 
 v 
ln 7(cos v)71/ v
 71/ v sin v 
v2
12. f ( x)  xe  e x  f ( x)  exe1  e x
23 x
x(ln 2)23 x (3)  23 x [(3ln 2) x  1]23 x

13. f ( x) 
 f ( x) 

x
x2
x2
14. y  2cot x  y  (ln 2)2cot x ( csc2 x)   ln 2(csc2 x)2cot x
15. f ( x)  log 2 ( x 2  x  1) 
ln( x 2  x  1)
1
2x 1
2x 1
 f ( x) 
 2
 2
ln 2
ln 2 x  x  1 ( x  x  1) ln 2
ln(t 2  1)
2t
t
 f (t ) 
 2
2
ln10
(2ln10)(t  1) (t  1) ln10
y
17. y  3x  ln y  ln 3x  x ln 3   ln 3  y  3x ln 3
y
16. f (t )  log t 2  1 
1
2
18.
y  ( x  2)1/ x  ln y  ln( x  2)1/ x 
1
y
1
1 1
ln( x  2)    2 ln( x  2)  

x
y
x
x x2
 1
ln( x  2) 
1/ x
y  

 ( x  2)
2
x
(
x

2)
x


19.
y  ( cos x ) x  ln y  ln( cos x ) x  x ln cos x 
y 
1
(cos x) 1/2 (  sin x)
y
 ln cos x  x  2

y
cos x
cos x ln(cos x)  x sin x
( cos x ) x
2 cos x
1
3x
2

20.  3 dx 
0
ln 3 0 ln 3
1
x
21. Let u  x 2  2 x , so du  2( x  1)dx . Then
3u
3x  2 x
(
x

1)3
dx

3
du


C

C .


2 ln 3
2 ln 3
22. Let u  2 x , so du  2 x ln 2dx . Then
2
x2  2 x
x
x
 2 sin 2 dx 
1
2
u
1
cos u
cos 2 x
sin
u
du



C


C
ln 2 
ln 2
ln 2
23
Chapter 7 The Transcendental Functions
23. Let u  1  3x , so du  3x ln 3dx . Then
3x
1 du ln u
ln(3x  1)
 1  3x dx  ln 3  u  ln 3  C  ln 3  C .
24.
I
I
a. 5  log  105   I  105 I 0
I0
I0
b. 8  log
I
 I  108 I 0 . Denote the intensities of the earthquakes with magnitudes
I0
I8 108 I 0
5 and 8 by I 5 and I 8 , respectively. Then
 5  103  I8  1000 I 5 . Thus, the
I 5 10 I 0
earthquake with magnitude 8 is 1000 times as intense as that with magnitude 5.
c. Denote the intensity of the Tangshan earthquake by IT and that of the San
Francisco earthquake by I S . Then IT  108.2 I 0 and I S  106.9 I 0 , so
IT 108.2 I 0
 6.9  101.3  IT  101.3 I s  20 I s Thus, the Tangshan earthquake was
I S 10 I 0
approximately 20 times as intense as the San Francisco earthquake.
25.
a. We solve 29.92e0.2 x  20  e0.2 x
20

x
29.92
ln
20
29.92  2.014 mi .
0.2
b. Differentiating p( x)  29.92e0.2 x with respect to t gives
dp
e0.2 x
dp
dx
dx
dt , so if x  2.014 and dp  1 , then
 29.92(0.2)e0.2 x


dt
dt
dt
dt
5.984
dx
e0.22.014 (1)

 0.25 and the balloon is rising at the rate of approximately 0.25
dt
5.984
mi/h.
26. y  2x  1  y  2x ln 2  m  2x ln 2
x 0
 ln 2 , so an equation of the tangent line
is y  2  ln 2( x  0) or y  (ln 2) x  2 .
log x
27. f ( x) 
 f ( x) 
x
x
1 1
  log x
1  ln x
ln10 x

 x  e . From
2
x
(ln10) x 2
the sign diagram, we see that f is increasing on (0, e) and decreasing on (e, ) .
28. A  
10
1
log x
1
dx . Let u  log x , so du 
dx , x  1  u  0 , and
x
x ln10
24
Chapter 7 The Transcendental Functions
1
x  10  u  1. Thus, A  ln10 u du 
0
1
1
0
0
1
ln10 2
ln10
.
u 
2
2
0
29. V  2  xy dy  2  x  3x dx . Let u  x 2 , so du  2x dx . Then
2
1
V  2 
1
1
0 2
3u
2
3   

.
ln 3 0 ln 3
u
30.
15
 $8052.55
a. A  5000(1  0.1
1 )
25
 $8144.47
b. A  5000(1  0.1
2 )
45
 $8193.08
c. A  5000(1  0.1
4 )
125
 $8226.54
d. A  5000(1  0.1
12 )
0.1 3655
)
 $8243.04
e. A  5000(1  365
f. A  5000e0.15  $8243.61
4.4
31. We solve the equation 4.4  2.1e6r  e6r 
r
2.1
ln
4.4
2.1  0.123 or 12.3%
6
per year.
32. He needs A  75,000e0.0510  $123,654.10 .
33.
a. Let f ( x)  e x  x  1 . Then f (0)  0 . Furthermore, f ( x)  e x  1  0 if x  0 . So
f is increasing on (0, ) and therefore, for x  0 , f ( x)  e x  x  1  0  e x  x  1. .
b. Let f ( x)  e x  1  x 
1 2
x . Then f ( x)  e x  1  x  e x  (1  x) . Using the result
2
25
Chapter 7 The Transcendental Functions
of part a, we see that f ( x)  0 for x  0 . Since f (0)  0 , we see that
f ( x)  e x  1  x 
1 2
1
x  f (0)  0 , if x  0 , that is e x  1  x  x 2 , for x  0 .
2
2
34. The integrand f ( x)  2cosx satisfies f ( x)  2cos(  x )  2cosx  f ( x) , and so f is an
even function. Therefore, 
1/2
1/2
35. True:
logb x

log a x
ln x
ln b
ln x
ln a

1/2
2cos x dx  2 2cos x dx .
0
ln a
.
ln b
36. True. If we let f ( x)  a x , then
a x 1
f ( x)  f (0)
d
 lim
 f (0)  (a x )  (ln a)a x
x 0
x 0
x
x0
dx
x 0
f (0)  lim
a x 1
lim
 ln a .
x 0
x
37. True. If we let f ( x)  logx , then
f (3)  lim
x 0
f (3  x)  f (3) d
1
1
.
 (log x) 

x
dx
x ln10 x 3 3ln10
x 3
x 0
 ln a , so
26
Chapter 7 The Transcendental Functions
7.5 Inverse Trigonometric Functions
1. sin 1 0  0
2. sin 1 12  6
3. tan 1 1  4
4. tan 1 3  3
5. sin 1
6. tan 1
3
2
 3
1
3
 6
7. sec 1 2  3
8. sin 1 ( 12 )  6
9. sin(sin 1
1
2
)  sin 4 
2
2
10. cos(sin 1
3
2
)  cos 3  12
11. tan(sin 1
2
2
)  tan 4  1
12. Let   sin 1 53 . Then sin   53 , so sec(sin 1 53 )  sec   54 .
13. Let   sin 1 x . Then sin   x , so cos(sin 1 )  cos   1  x 2 .
14. tan(tan 1 x)  x because tan and tan−1 are inverse functions.
27
Chapter 7 The Transcendental Functions
15. Let   sin 1 x . Then sin   x , so sec(sin 1 x)  sec  
1
1  x2
.
16. Let   tan 1 x . Then, tan   x so
sin(2 tan 1 x)  sin 2  2sin  cos   2 
x
1  x2

1
1  x2
17. f ( x) 
d
1
d
3
sin 1 3x 
 (3x) 
2
dx
1  (3 x) dx
1  9 x2
18. f ( x) 
d
1
d
2x
tan 1 x 2 
 ( x2 ) 
2 2
dx
1  ( x ) dx
1  x4
19. g (t ) 
20. f (u ) 

2x

1  x2
d
(3t )
d
3t
[t tan 1 3t ]  tan 1 3t  t  dt
 tan 1 3t 
2
dt
1  (3t )
1  9t 2
d
d
1
du (2u )
sec1 2u 

2
du
| 2u | (2u)  1 | u | 4u 2 1
21. h( x) 
d
1
2
1
(sin 1 x  2 cos 1 x) 


dx
1  x2
1  x2
1  x2
22. g ( x) 
d
1
1
1 x
(tan 1 x  x cot 1 x) 
 cot 1 x  x 

 cot 1 x
2
2
2
dx
1 x
1 x
1 x
23. y 
d
1
[( x 2  1) tan 1 x]  2 x tan 1 x  ( x 2  1)  2
 2 x tan 1 x  1
dx
x 1
( t 1)  ( t 1)
d t 1
2
( )
d
t 1
2
1
24. g (t )  tan 1
 dt tt 11 2  (t (1)t 1)2 
 2
2
2
dt
t  1 1  ( t 1 )
t 1
1  (t 1)2 (t  1)  (t  1)
25. y 
d
(sin 2 x)
d
2 cos 2 x
tan 1 (sin 2 x)  dx

2
dx
1  (sin 2 x) 1  sin 2 2 x
26. f ( x) 
d
1
d 2x
2e2 x
1
2x
sin (e ) 
e 
dx
1  (e2 x )2 dx
1  e4 x
27. Let   cos 1 x 2 . Then cos   x 2 , so
28
Chapter 7 The Transcendental Functions
x2
h( x)  cot(cos 1 x 2 )  cot  

1  x4
(1  x 4 )1/2 (2 x)  x 2 ( 12 )(1  x 4 )1/2 (4 x3 )
2x
.
h( x) 

4
1 x
(1  x 4 )3/2
28. f ( ) 
d
d
1
(sec1  )1  (sec1  ) 2
(sec1  )  
d
d
|  | (sec1  ) 2  2  1
29. f (t ) 
d 1
2
 1 8t
ln(1  4t 2 )  t tan 1 2t   
 tan 1 2t  t 
  tan 1 2t
2
2

dt  4
1  4t
 4 1  4t
30.
f ( x)  sin 1 x  2 x  f ( x) 
1
1 x
2
2 0 
1
1 x
2
 2  1  x2 
1
3
 x
,
4
2
the critical number of f.
From the sign diagram, we see that f has a relative minimum of f ( 23 )  3  3 and a
relative maximum of f ( 23 )  3  3 .
31.
1
f ( x)  sin 1 x  f ( x) 
 f ( x) 
d
x
(1  x 2 ) 1/2 
0 x 0.
dx
(1  x 2 )3/2
1 x
Because f ( x)  0 on (1, 0) and f ( x)  0 on (0,1) , we see that f has an
2
inflection point at (0, 0) .
32.
dx

Then
16  x 2


dx

dx
16[1  ( x / 4) 2 ]
du

1
dx
, Let u  14 x , so du  14 dx .

2
4 1  ( x / 4)
 sin 1 u  C  sin 1 ( 14 x)  C .
16  x
1 u
1/2
1/2
dx
dx
33. 
. Let u  2x , so du  2 dx , x  0  u  0 , and

2
0 1 4x
0 1  (2 x) 2
2
2
29
Chapter 7 The Transcendental Functions
1
x   u  1. Then 
1
2
1/2
0
dx
1 1 du
tan 1 u
1  


   .
2
2

1 4x
2 0 1 u
2 0 2 4  8
34. Let u  19 x 2 , so du  92 x dx . Then
dx
x
x 4  81

2
1
x dx
1 9
du
1
1
1
1  x 



sec
u

C

sec
 C
9  x 2 ( x 2 / 9) 2  1 9 2  9u u 2  1 18
18
 9 
1
dt
3t 2 dt
t3

 
35. 
. Let u  , so du 
.
4
4
t t 6  16
t 16[(t 6 /16)  1] 4 t (t 3 / 4) 2  1
dt
dt
Then
3
1
t 2 dt
1 4 1
du
1
1
1
1 | t |
 t t 6  16  4  t 3 (t 3 / 4)2  1  4  3  4  u u 2  1  12 sec | u |  C  12 sec 4  C
dt
1
e2 x
e2 x
2x
2x
dx

0 1  e4 x
0 1  (e2 x )2 dx . Let u  e , so du  2e dx , x  0  u  1 , and
1
36.
e2
e2 x
1 e2 du
1
1
x  1  u  e . Then 
dx


tan
u
 12 (tan 1 e2  4 ) .
2
2

1
0 1  e4 x
1
2 1 u
1
2
sin x
37.


sin x
4  cos x
2
dx 
1
sin x
dx . Let u  12 cos x , so du   12 sin x dx . Then

2 1  ( cos2 x ) 2
1
du
dx  (2) 
  sin 1 u  C   sin 1 ( cos2 x )  C .
2
2
4  cos x
1 u
2
38.
1
x3
x3
4
3
dx

0 1  x8
0 1  ( x4 )2 dx . Let u  x , so du  4 x dx , x  0  u  0 , and
1
x3
1 1 du
1 1
0 1  x8 dx  4 0 1  u 2  14 tan u 0  14 ( 4 )  16 .
dx
39. Let u  sin 1 x , so du 
, x  0  u  0 , and x  23  u  3 . Then
2
1 x
x  1  u  1. Then

0
3/2
sin 1 x
1  x2
dx  
 /3
0
1
u2
u du 
2
40. Let u  tan 1 x , so du 
 /3

0
2
18
.
dx
. Then
1  x2
tan 1 x
2
1
2
 1  x2 dx   u du  12 u  C  12 (tan x)  C .
41. Let u  x , so du  2 1 x dx . Then
30
Chapter 7 The Transcendental Functions
dx
1
dx
1
du
1
du
. Next, let v  12 u ,
 
  2
 
2
2
1
1
1
1  ( 4 u)
2 1  ( 2 u)
x (4  x) 4
x (1  4 x) 4
1
so dv  2 du . Then

dx
1
dv
  2
 tan 1 v  C  tan 1 ( 12 u )  C  tan 1 ( 2x )  C .
2
2
1

v
x (4  x)

42.
dx
 4  ( x  2)
dx
 4  ( x  2)
2
2

1
dx
. Let u  12 ( x  2) , so du  12 dx . Then

4 1  ( x 2 2 ) 2
1
du
  2
 12 tan 1 u  C  12 tan 1 ( x 2 2 )  C .
2
4
1 u
dx
1 1 dx
 
. Let u  12 x , so du  12 dx , x  0  u  0 , and
2
0 4 x
4 0 1  ( 12 x) 2
43. A  
1
1/2 du
1/2
1
 12 tan 1 u  12 tan 1 12 .
x  1  u  12 . Then A   2
2
0
4 0 1 u
44.
Let P( x, 0) . We have cot  
cot  
x
and
150
1000  x
, so
40
          cot 1
x
1000  x
 cot 1
for
150
40
0  x  1000 . Thus,
1
 401
   150x 2 
x 2
1  ( 150 ) 1  ( 1000
40 )
=
150
40

2
22,500  x 1600  1, 000, 000  2000 x  x 2

110 x 2  300, 000 x  149,340, 000
(22,500  x 2 )( x 2  2000 x  1, 001, 600)
Setting    0 gives 110 x 2  300,000 x  149,340,000 . Using the quadratic
formula, we find x  655 , the only critical number in (0,1000) . From the table, we
see that θ is maximized at x  655 , so the restaurant should be located
approximately 655 feet from the longer jetty.
45.
31
Chapter 7 The Transcendental Functions
Referring to the figure, we see that
t
  tan 1  10200t   tan 1  20
 , 0  t  15 .
2
d

dt 1 
1
10

2
t
1 2
20
t

2

40t

400  t 4
d 2  400  t  (40)  (40t )  4t

2
dt 2
 400  t 4 
4
3
  120t  16, 000  0 
 400  t 
4
4
2
120t 4  16,000  t  3.398 . From the sign diagram, we see that   is maximized
when t  3.398 , at which time the position of the car is approximately


d  10t 2  10 3.3982  115.5 ft from the starting point..
46. With L  10 and r  1 , we have V  10  12   sin 1 h  h(1  h 2 )1/2  .
Differentiating both sides of the equation with respect to t, we have
dh

dV
dh 
 10  dt  (1  h 2 )1/2  h( 12 )(1  h 2 ) 1/2 (2h)  
2
dt
dt 
 1 h

10
10h 2
2
 
 10 1  h 
2
1  h2
 1 h
Substituting h  0.4 and
0.2  20 1  (0.4) 2
 dh
dh
 20 1  h 2

dt
 dt
dV
 0.2 into the equation gives
dt
dh
dh
 0.011 ft / s.
, and solving, we find
dt
dt
47. False: sin 1 x  arcsinx is not equal to (sinx) 1 
1
 cscx.
sinx
48. False. Let x  0 . Then (sin 1 0)2  (cos1 0)2  0  ( 2 )2  4  1 .
2
49. True. f ( x) 
d
1
cos 1 x  
 0 for x  (1,1) , and so f is decreasing.
dx
1  x2
32
Chapter 7 The Transcendental Functions
7.6 Hyperbolic Functions
1.
a. sinh 2  12 (e2  e2 )  3.6269
b. cosh 4  12 (e4  e4 )  27.3082
c. sech3  0.0993
2.
a. cosh 0  1
b. sech(1)  0.6481
c. csch(ln 2)  1.3333
3.
a. csch 1 2  0.4812
b. csch 1 (2)  0.4812
c. coth 1 ( 32 )  0.8047
e x  e (  x ) e x  e x
4. cosh( x) 

 cosh x
2
2
2
2
 1   sinh x  1  sinh x cosh x



1
5. sech x  tanh x  
 

cosh 2 x
cosh 2 x
 cosh x   cosh x 
2
2
6.
2
2
1  cosh 2 x (cosh 2 x  sinh 2 x)  (cosh 2 x  sinh 2 x) 2cosh 2 x


 cosh 2 x
2
2
2
7.
2
2
 e x  e x   e x  e x 
e 2 x  2  e 2 x  e 2 x  2  e 2 x
cosh x  sinh x  


 

4
 2   2 
e 2 x  e 2 x

 cosh 2 x
2
2
2
8.
e x  e x e y  e y e x  e x e y  e y



2
2
2
2
x y
x y
x y
x  y
x y
e e e e e e e e
e e  e xe y  e xe y  e xe y


4
4
x y
x  y
x y
( x y )
2e e  2e e
e e
=

 cosh( x  y )
4
2
cosh x cosh y  sinh x sinh y 
9. sinh x  43 ,csch x  sinh1 x  43 ,cosh x  1  sinh 2 x  1  ( 34 )2  53 ,sech x  cosh1 x  53 ,
33
Chapter 7 The Transcendental Functions
sinh x
4
and coth x 
tanh x  cosh
x  5
1
tanh x
 54 .
10. f ( x) 
d
d
sinh 3x  cosh 3 x  (3 x)  3cosh 3 x
dx
dx
11. g ( x) 
d
d
tanh(1  3x)  sech 2 (1  3 x)  (1  3 x)  3sech 2 (1  3 x)
dx
dx
12.
d t
(e sinh t )  et sinh t  et cosh t  et (sinh t  cosh t )
dt
 et  e  t et  e  t  2 t
 et 

e
2 
 2
f (t ) 
13. F ( x) 
d
sinh x
ln(cosh x) 
 tanh x
dx
cosh x
14. g (u) 
d
2u sinh u 2
tanh(cosh u 2 )  sech 2 (cosh u 2 )  sinh u 2  2u 
du
cosh 2 (cosh u 2 )
15.
d
cosh 2 (3t 2  1)  2cosh(3t 2  1)sinh(3t 2  1)  6t
dt
 12t cosh(3t 2  1)sinh(3t 2  1)
f (t ) 
16. g (v) 
d
(v sinh v 2 )  sinh v 2  v cosh v 2  2v  sinh v 2  2v 2 cosh v 2
dv
17. f ( x) 
d
tanh(e 2 x  1)  sech 2 (e 2 x  1)  2e 2 x  2e 2 x sech 2 (e 2 x  1)
dx
18.
d
(cosh x  sinh x) 2/3  23 (cosh x  sinh x) 1/3 (sinh x  cosh x)
dx
  23 (cosh x  sinh x) 2/3
f ( x) 
19. g ( x) 
d
1
sinh x
tanh 1 (cosh x) 
 sinh x 
  csch x
2
dx
1  cosh x
1  cosh 2 x
20. f ( x) 
d sinh x
(1  cosh x) cosh x  (sinh x)sinh x
1


2
dx 1  cosh x
(1  cosh x)
1  cosh x
(1  tanh 2t )  12  2cosh 1 t sech 2 2t
d cosh 1 t
t 1
21. y 

dt 1  tanh 2t
(1  tanh 2t )2
d
3
sinh 1 3 x 
dx
1  9x2
d
1
1
1
cosh 1 2 x  (cosh 1 2 x) 1/2 
2 
23. y 
2

1
dx
2
4x 1
cosh 2 x 4 x 2  1
d
1
1
24. f ( x)  sech 1 2 x  1  

dx
(2 x  1) 1  (2 x  1)
(2 x  1) 2 x
22. f ( x) 
34
Chapter 7 The Transcendental Functions
d
2x
2x2
( x cosh 1 x 2 )  cosh 1 x 2  x 
 cosh 1 x 2 
dx
x4 1
x4 1
d
1
d
1
26. f ( x)  sech 1 x  
 x1/2  
dx
x 1  x dx
2x 1  x
25. y 
d
9x
3
9( x  1)
( 9 x 2  1  3cosh 1 3 x) 
 3

2
2
dx
9x 1
9x 1
9x2 1
28. Let u  2 x  3 , so du  2 dx . Then
27. y 
 cosh(2 x  3) dx   cosh u du 
1
2
1
2
sinh u  C  12 sinh(2 x  3)  C .
29. Let u  sinh x , so du  cosh dx . Then

sinh x cosh x dx   u1/2 du  23 u 3/2  C  23 (sinh x)3/2  C
30. u  3x  du  3 dx , so
1
1 cosh u
 coth 3x dx  3  coth u du  3  sinh u du 
1
3
ln | sinh u | C  13 ln | sinh 3 x | C .
31. Let u  1  cosh x , so du  sinh x dx . Then
sinh x
 1  cosh x dx  
du
 ln | u | C  ln(1  cosh x)  C .
u
32.
2
b
b
b
x
x
V    y 2 dx  2   a cosh  dx  2 a 2  cosh 2 dx
b
0
0
a
a

b
=2 a
2

b
0
1
2x 
a
2x 
2
1  cosh  dx   a  x  sinh 
2
a 
2
a 0

a
2b 
2b 

b 1
= a 2  b  sinh    a 3   sinh 
2
a 
a 

a 2
33.
d
d  eu  e  u
cosh u 

dx
du  2
 du eu  eu du
du

 sinh u

2
dx
dx
 dx
35
Chapter 7 The Transcendental Functions
34.
d
d  1  du  sinh u du
du
sech u 

  tanh u sech u


2
dx
du  cosh u  dx cosh u dx
dx
3
 e x  e x e x  e x 
3x

35. True: (sinh x  cosh x)  
  e  0 for all x.
2 
 2
3
36. True. The integrand f ( x)  cosxsinhx is odd because

f ( x)  cos( x)sinh( x)  cosx(sinhx)   f ( x) , so  cos x sinh x dx  0

36
Chapter 7 The Transcendental Functions
7.7 Indeterminate Forms and l’Hôpital’s Rule
1. lim
x 1
1 1
 lim

2
x  1 x1 2 x 2
2. lim
x3  8
3x 2
 lim
 12
x  2 x 2 1
x 1
x 2
ex 1
ex

lim
1
x 0 x 2  x
x 0 2 x  1
sin t
cos t
 lim
1
4. lim
t    t
t  1
3. lim
2sec2 2
2
 0
 0

1
x  cos x
1  sin x
 lim
6. lim
. Observe that the limit on the right-hand side does not
x  2 x  1
x 
2
5. lim
tan 2
 lim
exist, so l’Hôpital’s Rule does not apply. We evaluate the limit as follows:
1  cosx x 1
x  cos x
lim
 lim
 .
x  2 x  1
x  2  1
2
x
sin x  x cos x
x sin x
1
x
1
 lim
 lim
cos 4 x 
3
2
2
x 0
x 0 3 tan x sec x
tan x
3 x 0 sin x
3
7. lim
8. lim
x 
x
1/ (2 x )
x
 lim
 lim

x

x

ln x
1/ x
2
9.
3(ln x) 2 ( 1x )
6 ln x( 1x )
(ln x)3
3(ln x) 2
3ln x
lim

lim

lim

lim
 lim
2
2
x 
x

x

x

x

x
2x
2x
4x
2x2
3/ x
3
 lim
 lim 2  0
x  4 x
x  4 x
ln(1  e x )
e x / (1  e x )
1

lim
 lim
0
2
x
x 
x 
x  (1/ e  1)(2 x)
x
2x
10. lim
11. lim
x 1 3
1
( x  2) 1/2  1 12  1 9
x2x
 lim 2 2
 2 
2/3
4
2 x  1  1 x1 3 (2 x  1)
3
ex  x 1
2 xe x  1
 lim
 1  x2  
12. lim
2
x 0
x 0
x
1 1 x
2
2
sin x  x
cos x  1
 sin x
cos x
1
 lim x  x
 lim x  x   lim x  x  

x
x 0 e  e
x 0 e  e
 2 x x 0 e  e  2 x 0 e  e
2
13. lim
x
sin 1 (2 x)
2 / 1  4 x2
 lim
2
x 0
x 0
x
1
14. lim
37
Chapter 7 The Transcendental Functions
15.
1
1
x  tan x
1  sec 2 x

 1
lim  cot x    lim 
   lim
 lim
x 0
x  0 tan x  x sec 2 x
x  x 0  tan x x  x 0 x tan x

2sec 2 x tan x
 lim
0
x  0 2sec 2 x  2 x sec x tan x
(sin x)2
2sin x cos x
2sin x cos x
  lim
  lim
  lim 2cos3 x  2
x 0 1  sec x
x 0 sec x tan x
x 0
x 0
sin x
2
cos x
16. lim
sinh u
cosh u
 lim
1
u  0 sin u
u 0 cos u
17. lim
1
1  cos x  x
sin x  1
1

 lim
 lim
 
18. lim  

x 0  x
1  cos x  x 0 x(1  cos x) x 0 1  cos x  x sin x
19.
1
1
1
1
x

1

ln
x
x 1


x
lim 

 lim
 lim
 lim

x 1 ln x
x  1  x 1 ( x  1) ln x x 1 ln x  x  1 x 1 x ln x  x  1

x
1
1
 lim

x 1
2
1
ln x     1
 x
 cos x
ln(1  sin x)
1
20. lim[csc
x ln(1  sin x)]  lim
 lim 1  sin x  lim
 1

x 0
x 0
x 0
x 0 sin x  1
sin x
cos x
1
1

21. lim  e  x   lim x  0
x  x

 x  xe
22. lim(1
 cos x)tan x is an indeterminate form of type 00 , so let y  (1  cosx) tan x .

x 0
Then ln y  ln(1  cos x) tan x  tan x ln(1  cos x) , so
sin x
ln(1  cos x)
ln lim y  lim (ln y )  lim tan x ln(1  cos x)  lim
 lim 1  cos2 x
x 0
x 0
x 0
x 0
x 0  csc x
cot x
3
2
sin x
3sin x cos x
= lim
 lim
 lim 3sin x cos x  0
x 0 1  cos x
x 0
x 0
sin x


Thus, lim y  lim(1
 cos x)tan x  e0  1 .

x0
x0
23. lim(ln x)1/ x is an indeterminate form of type 0 . Let y  (ln x)1/ x . Then
x 
38
Chapter 7 The Transcendental Functions
lny  ln(lnx)1/ x 


1
ln ln x , so
x
1/ x
ln ln x
1
 lim ln x  lim
 0 , and thus
x 
x  1
x  x ln x
x
ln lim y  lim(ln y)  lim
x 
x 
lim y  lim(ln x)1/ x  e0  1 .
x
x
1
24. lim  
x 0  x 
tanh x
1
is an indeterminate form of type 0 , so let y   
 x
1
Then ln y  ln  
x


tanh x
 tanh x ln
ln lim y  lim (ln y)  lim
x 0
x 0
x 0
1
, and so lim y  lim  
x 0
x 0  x 
25.
tanh x
1
  tanh x ln x . We calculate
x
 ln x
1/ x
sinh 2 x
2sinh x cosh x
  lim

lim
 lim
0
2

x 0  csch x
x 0
x 0
coth x
x
1
tanh x
 e0  1 .
lim  (tan x)cos x is an indeterminate form of type 0 , so let y  (tan x)cos x .
x  ( /2)
Then ln y  ln(tan x)cos x  cos x ln tan x . We calculate
sec2 x
ln(tan x)
cos x
ln lim  y  lim  ln y  lim 
 lim  tan x  lim 
0,
2
x ( /2)
x ( /2)
x ( /2)
x (  /2) sec x tan x
x (  /2) sin x
sec x


lim  y  lim  (tan x)cos x  e0  1 .
so
x ( /2)
x ( /2)
26. lim(1  1/ x) x  lim[(1  1/ x) x ]x   . Compare with Example 10, in which we
3
2
x 
x 
calculated lim(1  1/ x) x  e .
x 
 2x 1 
27. lim 

x  2 x  1


 2x 1 
ln y  ln 

 2x 1 


x
 2x 1 
is an indeterminate form of type 1 . Let y  
 , so
 2x 1 
x

x
 x ln
4
ln 22 xx 11
8( 32 x1/2 )
8 x3/2
3
1 4 x 2

lim

lim

lim

lim
0

1/2

3/2
2
x  x
x   1 x
x  4 x  1
x 
x  2 x1/2
8x
2
ln lim y  lim ln y  lim
x 
x 
2x 1
. Then
2x 1
39
Chapter 7 The Transcendental Functions
 2x  1 
, so lim y  lim 

x 
x  2 x  1


x
 e0  1 .
28. lim( x  x  1)  lim
2
x 
( x  x 2  1)( x  x 2  1)
( x  x  1)
x 
 lim
x 
2
1
x  x2  1
0
29.
lim
x 
2 tan 1 x  
e1/ x  1
2
30. lim
x 0
 lim
x 
2 / (1  x 2 )
e1/ x (2 / x3 )
2
  lim
x 
x3
(1  x 2 )e1/ x
2
  lim
x 
x
(1/ x 2  1)e1/ x
2
 
1
ln x
1
sin x
1
x
 lim cos
 lim
cos x 
x
2  3ln(sin x) x0 3( sin x ) 3 x0 x
3
x5  1
5x4 5
5x 4
.
Because
is not an indeterminate form, we

lim

lim
x 1 x 2  1
x 1 2 x
x 1 2x
2
cannot use l’Hôpital’s Rule.
31. lim
mt
mt
r
r
r



32. Let y  1   . Then ln y  ln 1    mt ln 1   , so
 m
 m
 m
r

r / m2
ln 1  
r
r
m

ln lim y  lim ln y  lim mt ln 1    t lim 
 t lim 1  r / m  t lim
 rt
m 
m 
m 
m

m

m

1
1
r
 m
 2
1
m
m
m


mt
r

. Therefore, lim y  lim 1    e rt , so
m 
m 
 m
mt
mt
r
r


A  lim P 1    P lim 1    Pe rt .
m 
m 
 m
 m
t  Rt / L
e
V
1  e Rt / L
Vt
(1  e Rt / L )  V lim
 V lim L

R 0
R 0 R
R 0
R 0
R
1
L
34. Substituting a  b into the expression for x gives an indeterminate form of type
33. lim I (t )  lim
0 / 0 . We use l’Hôpital’s Rule:
lim x  lim
a b
a b
= lim
a b
ab 1  e(b  a ) kt 
a  be
( b  a ) kt
ab 1  e


b 1  e (b  a ) kt    ab(1)(kt )e (b  a ) kt 
 lim
 lim
( b  a ) kt
d
a b
a b
1  b(kt )e(b  a ) kt


da  a  be
b 1  (1  akt )e(b  a ) kt 
1  bkte(b  a ) kt
d
da

( b  a ) kt
b 1  (1  bkt ) 
1  bkt

b 2 kt
a 2 kt
or
1  bkt
1  akt
35. Since k  0 , lim x k   , and so we can apply l’Hôpital’s Rule to obtain
x 
40
Chapter 7 The Transcendental Functions
ln x
1/ x
1
 lim k 1  lim k  0 .
k
x  x
x  kx
x  kx
lim
1
1
x sin
x  lim
x . Since  | x | x sin 1 | x | , the Squeeze Theorem
36. lim
x 0 sin x
x 0 sin x
x
x
x 2 sin
shows that lim x sin
x 0
1
 0.
x
1
1  1 

1
1
1
2 x sin  x 2  cos   2 
2 x sin  cos
x
x  x 

x  lim
x
x . Since the limit
lim
 lim
x 0 sin x
x 0
x

0
cos x
cos x
x 2 sin
in the numerator does not exist, l’Hôpital’s Rule is not applicable here.
37.
38.

lim
x

lim
x
0
x 0
x
0
x 0
cos t 2 dt
sin t 2 dt
x3
 lim
d
dt

0
x 0
 lim
x 0
x
cos t 2 dt
d
dx
d
dt

x
0
d
dx
( x)
sin t 2 dt
( x3 )
cos x 2
1
x 0
1
 lim
sin x 2 1
sin x 2 1

lim

x 0 3 x 2
3 x 0 x 2
3
 lim
39. False. Let f ( x)  x3 and g ( x)  x 2 . Then lim f ( x)  0 and lim g ( x)  0 .
x 0
lim
x 0
x 0
f ( x)
x3
 lim 2  lim x  0 . On the other hand,
x 0
g ( x) x0 x
f ( x)
d f ( x)
d f ( x)
d x3
d
.
 lim
 lim
 lim ( x)  lim1  1 , so lim
2
x 0 g ( x )
x 0 dx g ( x )
x 0 dx g ( x)
x 0 dx x
x 0 dx
x 0
lim
41
Chapter 7 The Transcendental Functions
Chapter 7 Review
1. ln x  52  x  e 2/5
2. log3 x  2  x  32  9
3. e
x
 4  ln e
x
 ln 4  x ln e  ln 4  x  ln 4  x  (ln 4)2
4. 2  3e x  6  3e x  4  e x  43  ln e x  ln 34   x  ln 34  x  ln 34
5. ln x  ln( x  2)  0  ln x( x  2)  0  x( x  2)  1  x 2  2 x  1  0
2 44
 1  2 . We reject the negative root, so the solution is x  1  2 .
2
6. 32 x  12  3x  27  0  (3x )2  12(3x )  27  (3x  3)(3x  9)  0  x  1 or x  2
7. tan 1 x  1  tan(tan 1 x)  tan1  x  tan1
x
8. y  e2 x  2  e2 x  y  2  ln e 2 x  ln( y  2)  2 x  ln( y  2)  x  12 ln( y  2)
9.
 y1/2 
y1/2
3
ln( x3 y / z 2 )  ln  x3

ln
x

ln
 ln x3  ln y1/2  ln z  3ln x  12 ln y  ln z

z
z


10.
2ln x  ln( x3 / y 2 )  4ln( x  y)1/2  ln x 2  ln
 ln
 2 x3
x3
1 
4(1/2)

ln(
x

y
)

ln
x  2 

2
y
y ( x  y)2 

x5
y 2 ( x  y)2
11. 3ln x  13 ln( yz )  6 ln( xy )1/2  ln x3  ln( yz ) 1/3  ln( xy )3  ln
x6 y3
x 6 y 8/3
 ln 1/3
3 yz
z
12. y 
d
d
1 d
1 1
1
ln x  1  ln( x  1)1/2 
ln( x  1)  

dx
dx
2 dx
2 x  1 2( x  1)
13. y 
d 1/2
x1/2 ln x  2
( x ln x)  12 x 1/2 ln x 

dx
x
2 x
14.
y 
15.
d x
[e (cos 2 x  3sin 2 x)]  e  x (cos 2 x  3sin 2 x)  e  x ( 2sin 2 x  6 cos 2 x)  5e  x (cos 2 x  sin 2 x )
dx
42
Chapter 7 The Transcendental Functions
 y 
1
x ln y  y ln x  3  ln y  x    y  ln x  y    0
x
 y
x

y ( y  x ln y )
y

 y   ln x      ln y   y  
x( x  y ln x)
x

y

16. y  ln( x 2 e2 x )  2 ln x  2 x  y 
17. y 
2
2(1  x)
2 
x
x
d x
[e (1  e  x ) 1/2 ]  e x (1  e  x ) 1/2  e x ( 12 )(1  e  x ) 3/2 (e x )
dx
=  12 (1  e x )3/2 [2e x (1  e x )  1] 
18. y 
2e x  3
2(1  e x )3/2
d csc x
d
(e )  ecsc x csc x   csc x cot ecsc x
dx
dx
19. y  ee  y  ee
x
x
x
x
d x
(e )  e e  e x  e e  x
dx
20. ye  xe  8  ye x  ye x  e y  xe y (2 yy)  0  y 
x
21. y 
y2
2
ye x  e y
2
2
e x  2 xye y
2
d x cot x
d
3
 3x cot x ln 3  ( x cot x)  3x cot x (cot x  x csc 2 x) ln 3
dx
dx
d
(cosh 2 x)(2)
[ln(sinh 2 x)] 
 2 coth 2 x
dx
sinh 2 x
d
1
1
23. y  ( x sec1 x)  sec1 x  x 
 sec1 x 
dx
| x | x2 1
| x | x2 1
22. y 
24. y 
d
2
2sec2 (cos 1 2 x)
[tan(cos 1 2 x)]  sec2 (cos 1 2 x) 

dx
1  4 x2
1  4 x2
25.
y 
d  1  x  1  
sin 
 
dx 
 x  2 
1
 x 1 
1 

 x2
2

d  x 1 
( x  2) 2
1
1





2
dx  x  2 
2 x  3 ( x  2)
| x  2 | 2x  3
26.
x cosh y  esinh y  10  cosh y  ( x sinh y ) y  esinh y (cosh y ) y  0
 y  
cosh y
(cosh y )esinh y  x sinh y
27. y 
d ax
e cosh bx  aeax cosh bx  beax sinh bx  e ax (a cosh bx  b sinh bx)
dx
43
Chapter 7 The Transcendental Functions
28. f ( x)  2 x 2  ln x  f ( x)  4 x 
number
1
2
(since x  0 ). From the sign diagram, we see that f is decreasing on
and increasing on
29. y 
1 4 x2 1
1

 0  x   , so f has a critical
x
x
2
 0, 12 
 12 ,  .
2x
2ln x  2 x(1/ x) 2(ln x  1)
. The required slope is
 y 

ln x
(ln x) 2
(ln x) 2
m  y xe  0 , so an equation of the tangent line is y  2e .
ln x
x(1/ x)  ln x 1  ln x
 f ( x) 

 0  x  e . f (1)  0 ,
x
x2
x2
f (e)  1/ e  0.368 , and f (5)  (ln 5) / 5  0.322 , so the absolute minimum value is
30. f ( x) 
f (1)  0 and the absolute maximum value is f (e)  1/ e .
31.
f ( x)  xlnx
(1) Since ln x is defined for x  0 , we see that
the domain of f is (0, ) . (2) Setting y  0
gives xlnx  0  lnx  0  x  1 , so the
x-intercept is 1.
(3) There is no symmetry.
(10)
(4) lim xlnx  
x 
(5) There is no asymptote.
(6) f ( x) 
d
( x ln x)  x( 1x )  lnx  1  lnx is continuous for x  0 and
dx
f ( x)  0  1  lnx  0  lnx  1  x  e 1  1e . From the sign diagram, we see
that f is decreasing on (0, 1e ) and increasing on ( 1e , ) . (7) From the sign
diagram, we see that f has a relative minimum at ( 1e ,  1e )  (0.368, 0.368) .
(8) f ( x)  1x is continuous on (0, ) and has no zero. Since f ( x)  0 for all
x  0 , we see that the graph of f is concave upward on (0, ) .
(9) f has no inflection point.
44
Chapter 7 The Transcendental Functions
32.
3
1  e x
(1) The domain of f is (, ) . (2) Setting
x  0 gives 32 as the y-intercept. (3) There is
f ( x) 
(10)
no symmetry. (4) lim f ( x)  0 and
x 
lim f ( x)  3 .
(5) From (3), we see that y  0
x 
and y  3 are horizontal asymptotes of f .
(6) f ( x)  3 dxd (1  e x ) 1
3e x
 0 for all x,
(1  e x )2
so f is increasing on (, ) .
 3(1  e x )2 (e x )(1) 
(7) f has no relative extremum.
(8) f ( x) 
(1  e x )2 (3e x )  3e x (2)(1  e x )(e x )
(1  e x )4
3e x (1  e x )[2e x  (1  e x )] 3e x (e x  1)
, so f ( x)  0  e x  1  x  0 .

(1  e x )4
(1  e x )3
From the sign diagram, we see that f is concave upward on (, 0) and concave
downward on (0, ) . (9) f has an inflection point at (0, 32 ) .

33. Let u  5x  3 , so du  5 dx . Then
1
1 du 1
 ln | u | C  15 ln | 5 x  3 | C .
u 5
 5x  3 dx  5 
34.
2

2
1
2
 x2
x3  2 x  1
2
1
2 
dx    x   x  dx    2 ln x  
2
1
x
x
x 1


 2
 (2  2 ln 2  12 )  ( 12  1)  2(1  ln 2)
35.
2x 1
2
1/ 3 
 3x  2 dx    3  3x  2  dx 
2
3
x  19 ln | 3 x  2 |  C
2
1
2t
 2u  C 
C.
36. Let u  t , so du  2t dt . Then  t  2 dt   2 du 
2 ln 2
2 ln 2
dx
37. Let u  ln x , so du 
. Then
x
2
t2
1
2
u
45
Chapter 7 The Transcendental Functions

sin ln x
dx   sin u du   cos u  C   cos ln x  C .
x
38. Let u  1  e3 x , so du  3e3 x dx , x  0  u  2 , and x  1  u  1  e3 . Then
1 e3
1
1 e3  1
.
 ln(1  e3 )  ln 2   ln
3
3
2
e3 x
1 1e3 du 1
dx

 ln | u |
0 1  e3x
3 2 u 3
2
1
39. Let u  sin 1 x , so du 

sin 1 x
dx
1  x2
. Then
dx   u du  12 u 2  C  12 (sin 1 x) 2  C .
1 x
40. Let u  1  sec t , so du  sec t tan t dt . Then
2
sec t tan t
 1  sec t
dt  
du
 ln | u | C  ln |1  sec t | C .
u
41. Because 1  e4 x  e4 x , we have
1
1
0
0
1  e4 x  e4 x  e2 x   1  e4 x dx   e2 x dx  12 e2 x  12 (e2  1) .
42. Using Part 1 of the Fundamental Theorem of Calculus along with the Chain Rule,
2
we calculate f ( x) 
4
2
d x2 et dt
ex
d 2
2 xe x
.

(x )  4
dx 0 t 2  1 ( x 2 )2  1 dx
x 1
43. A   xe x dx . Let u   x 2 , so du  2x dx , x  0  u  0 , and
2
0
x  4  u  16 . Then A   12 
16
0
eu du   12 eu
16
0
 12 (1  e16 ) .
2 ln x
dx
1 2 ln x 2
44. f av 
, x 1 u  0 ,
dx

2
dx . Let u  ln x , so du 


1
1
x
2 1
x
x
2 ln x
ln 2
ln 2
dx  2  u du  u 2  (ln 2) 2 .
and x  2  u  ln 2 . Then f av  2
0
1
0
x
x3  2 x 2  x
3x 2  4 x  1 0
45. lim
 lim
 0
x 1
x 1
x5  1
5x4
5
sin 2 x
2 cos 2 x 2
 lim

46. lim
x 0 sin 3 x
x 0 3cos 3 x
3
cos x
0
x  e x
47. lim e  x cos x  lim
x 
48.
1
x  sin x
1  cos x
 1
lim(csc x  1/ x)  lim 
   lim
 lim
x 0
x 0 sin x
x 0 sin x  x cos x
x  x0 x sin x

sin x
 lim
0
x 0 cos x  cos x  x sin x
46
Chapter 7 The Transcendental Functions
49.
1 
ex 1  x
ex 1
ex
1
1
1
lim   x   lim

lim

lim
 lim

x
x
x
x
x
x


x 0  x
x

0
x

0
x

0
x

0
e 1 
x(e  1)
e  1  xe
e  e  xe
2 x 2
50.
 x 1  x 1 x 1  x 1 
2
lim( x  1  x  1)  lim 

 lim
0


x 
x 
1
x  1  x  1  x x  1  x  1

sin x  x
cos x  1
 sin x
1
1
 lim
 lim
 lim cos3 x 
2
2
x 0 x  tan x
x 0 1  sec x
x 0 2sec x tan x
2 x 0
2
51. lim
52.
x(t )  aet  be t  v(t ) 

d
d
(aet  be t )  aet  be  t  a (t )  v(t )
dt
dt
d
(aet  be  t )  aet  be t
dt
53. y  200(e0.01x  e0.02 x )  y  2(e0.01x  2e0.02 x )  e0.01x  2e0.02 x  e0.03 x  2
1
 x  0.03
ln 2 . The absolute minimum value of y is y |x (ln 2)/0.03  378 , so the plane
gets to about 378 ft above the ground.